CODEWITHSUNDEEP
jqueryfadeinfadeout
technopeasant
technopeasant

Reputation: 7939

jQuery fade out then fade in

There's a bunch on this topic, but I havn't found an instance that applies well to my situation.

Fade a picture out and then fade another picture in. Instead, I'm running into an issue where the first fades out and immediately (before the animation is finished) the next fades in.

I read about this once and can't remember exactly what the trick was..

http://jsfiddle.net/danielredwood/gBw9j/

thanks for your help!

Upvotes: 37

Views: 118576

Answers (4)

user578895
user578895

Reputation:

fade the other in in the callback of fadeout, which runs when fadeout is done. Using your code:

$('#two, #three').hide();
$('.slide').click(function(){
    var $this = $(this);
    $this.fadeOut(function(){ $this.next().fadeIn(); });
});

alternatively, you can just "pause" the chain, but you need to specify for how long:

$(this).fadeOut().next().delay(500).fadeIn();

Upvotes: 84

Flimm
Flimm

Reputation: 150553

With async functions and promises, it now can work as simply as this:

async function foobar() {
  await $("#example").fadeOut().promise();
  doSomethingElse();
  await $("#example").fadeIn().promise();
}

Upvotes: 0

rabbit.aaron
rabbit.aaron

Reputation: 2579

After jQuery 1.6, using promise seems like a better option.

var $div1 = $('#div1');
var fadeOutDone = $div1.fadeOut().promise();
// do your logic here, e.g.fetch your 2nd image url
$.get('secondimageinfo.json').done(function(data){
  fadeoOutDone.then(function(){
    $div1.html('<img src="' + data.secondImgUrl + '" alt="'data.secondImgAlt'">');
    $div1.fadeIn();
  });
});

Upvotes: 5

user2000962
user2000962

Reputation:

This might help: http://jsfiddle.net/danielredwood/gBw9j/
Basically $(this).fadeOut().next().fadeIn(); is what you require

Upvotes: 2

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